Chris Khan 2007
 A saturated solution contains the maximum amount of solute that will dissolve in a specific solvent at a specific temperature.
 An unsaturated solution contains less solute than it has the capacity to dissolve.
 A supersaturated solution contains more solute than is present in a saturated solution.
 Crystallization is the process in which dissolved solute comes out of solution and forms crystals.
 The ease with which a solute particle replaces a solvent molecule depends on the relative strengths of three types of interactions:
o Solvent-solvent interaction
o Solute-solute interaction
o Solvent-solute interaction
 If this interaction is stronger than the solvent-solvent attraction and the solute-solute attraction, then the solution
process is favorable, or exothermic (∆Hsoln < 0).
 If this interaction is weaker than the solvent-solvent attraction and the solute-solute interaction, then the solution
process is endothermic (∆Hsoln > 0).
 Two liquids are said to be miscible is they are completely soluble in each other in all proportions.
 Solvation is the process in which an ion or a molecule is surrounded by solvent molecules arranged in a specific manner.
 Percent by mass is the ratio of the mass of a solute to the mass of the solution.
o Percent by mass of solute = mass of solute / (mass of solute + mass of solvent) x 100%
o 6.44g Naphthalene is dissolved in 80.1g Benzene. What is the percent by mass of Naphthalene? Do 6.44 / (6.44 + 80.1)
= about 7.44%.
 Mole Fraction (X) of A is written XA and is defined as moles of A / ∑ moles of all components.
 Molarity (M) is defined as moles of solute / liters of solution.
 Molality (m) is the number of moles of solute dissolved in 1kg of solvent.
o Molality = moles of solute / mass of solvent (kg)
o Calculate the molality of a solution containing 7.78g Urea in 203g Water. Do 7.78g / 60.062 g/mol to get 0.13 moles of
Urea. Divide this by 0.203 kg and get 0.638 m.
o Calculate the molality of a 5.86 M (C2H5OH) Ethanol solution with a 0.927g/mL density. First, we have to find that
there are 927 grams of Ethanol per liter of solution because the density is 0.927g/mL. Next, we have to find how much
water is in the solution by doing 5.86 mol x 46g/mol = 269.56g. Now, subtract 269.56g from 927g and get 657.44g, or
0.65744kg. Now, we can find the molality by dividing 5.86 moles by 0.65744 kg and get 8.9 m.
o Calculate the molality of a 44.6% (by mass) aqueous solution of NaCl. First, since NaCl is 44.6%, we can assume this is
44.6g out of 100g of solution, leaving 55.4g of water. If we convert the 44.6g NaCl to moles, we get 0.76mol. We can
divide this by 0.0554 kg of water to get about 13.8 m.
 Fractional Crystallization is the separation of a mixture of substances into pure components on the basis of their different
solubilities.
 Henry’s Law explains the relationship between gas solubility and pressure.
o The solubility of gas in a liquid is proportional to the pressure of the gas over the solution.
o C = kP where C is the molar concentration of the dissolved gas, P is the pressure (in atm) of the gas over the solution,
and k is a constant that depends only on temperature.
o Calculate the molar concentration of oxygen in water at 25 degrees Celsius for a partial pressure of 0.22 atm. The
Henry’s Law constant for oxygen is 1.3 x 10 -3 mol / L atm. Here, just use C=kP, which comes to C = (1.3 x 10 -3 mol / L
atm)(0.22 atm) = 2.86 x 10-4 mol / L.
 Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of solute
particles.
 Nonvolatile solutions do not have measureable vapor pressures. In these solutions, the vapor pressure of its solution is always
less than that of the pure solvent.
 Raoult’s Law states that the partial pressure of a solvent over a solution, P 1, is given by the vapor pressure of the pure solvent,
P1o, times the mole fraction of the solvent in the solution, X 1: P1 = X1P1o.
o Let’s say we want to calculate the VP of a solution made by dissolving 82.4g Urea in 212mL water at 35 C. What is that
VP and what is the VP lowering? First, convert Urea and water to moles. We get 11.78mol Urea and 1.37mol Water.
Now, we have to find the mole fraction of Urea, which is 11.78 / (11.78 + 1.37) = 0.896. Now, we have to find the VP
by using P1 = X1P1o. P1 = 0.896 x 42.19 = 37.79 mmHg. Lastly, to find the VP lowering, subtract 37.79 from 42.19 and
get 4.4 mmHg.
 Volatile solutions have measureable vapor pressure, which is the sum of the individual partial pressures. P A = XAPAo and PB =
XBPBo where PA and PB are the partial pressures over the solution for components A and B. P Ao and PBo are the vapor pressures of
the pure substances. XA and XB are their mole fractions.
 Ideal solutions are any solutions that obey Raoult’s Law. In an ideal solution, ∆Hsoln = 0.
 Fractional distillation is a procedure for separating liquid components of a solution based on their different boiling points. The
boiling-point elevation (∆Tb) is the boiling point of the solution (T b) minus the boiling point of the pure solvent (Tbo). ∆Tb =
Kbm where m is molality of the solution and Kb is the molal boiling-point elevation constant. Kb is in units of C / m. The
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freezing point depression (∆Tf) is the freezing point of the pure solvent (T fo) minus the freezing point of the solution (Tf). ∆Tf =
Kfm where m is the concentration of the solute in molality units, and K f is the molal freezing-point depression constant.
o Calculate the boiling and freezing points of a solution containing 478g of ethylene glycol in 3202g of water. First, covert
these to moles. We have 7.7mol EG and 3.2kg water. Now, we can divide these to find that the molality is 2.41 mol EG
/ kg H2O. Now, we can multiply this by 0.52 C / m in the equation ∆Tb = Kbm to get ∆Tb as 1.25 C. So, we add this to
100 C (boiling point) and get that the boiling point of this solution is at 101.25 C. To find the freezing point, we simply
use the equation ∆Tf = Kfm to find the ∆Tf = 1.86 m x 2.41 C / m to get ∆Tf as 4.48 C, subtracting that from 0 C to get
that the freezing point of this solution is -4.48 C.
Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated
one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. The
osmotic pressure of a solution is the pressure required to stop osmosis. π = MRT where pi is the osmotic pressure, M is the
solution’s molarity, R is the gas constant (0.082057 L atm / mol K), and T is the absolute temperature. Two solutions of equal
concentration are isotonic. The more concentrated solution is hypertonic and the dilute one is hypotonic. Crenation causes the
cell to shrink and to cease functioning. Transpiration is when the solute concentrations in leaf fluids increase.
o What is the osmotic pressure of a 0.884 M Urea solution at 16 C? Well, since π = MRT, we have to plug in the values
we know. π = (0.884 mol / L)(0.082057 L atm / mol K)(16+273.15)K = 20.97 atm.
Molality can also be found, not only as moles / kg, but as ∆T f / Kf.
o What is the molality of a solution of 0.85 g in 100.0 g of Benzene, which together have a 5.16 C freezing point? What is
the molar mass? First, we have to find the freezing point of benzene normally, which is 5.5 C. Now, we subtract 5.16
from this and divide this by K f since m = ∆Tf/Kf and we just found ∆T f. So, don’t forget about significant figures and put
only 0.3 C in the top. Divide this by 5.12 C /m and get the molality as 0.06 mol solute / kg solution. Now, we can find
the molar mass by first finding the number of moles. Since there are 0.06 moles of solute per kg, there are 0.006 moles
of solute per .1000 kg of benzene. We can divide 0.85 g by this and get 141 grams per mole. Unfortunately, we have to
put this to 1 significant figure and leave it at 100 grams per mole.
o Calculate the molar mass of a polymer whose osmotic pressure is 8.63 mmHg at 21 C when it is contained in a benzene
solution of 202mL and 2.47g of the polymer. π = MRT, so we can say that M = π/RT = (0.011 atm) / (0.082057 L atm /
mol K)(294.15K) = 4.7 x 10-4 mol / L or 4.7 x 10-7 mol / mL. Now, we can multiply this by 202 mL to get 9.5 x 10 -5
mol. Since we know we have 2.47 g of this stuff, we divide that by 9.5 x 10 -5 mol to get 26,000g/mol, or 2.6 x 10 4
g/mol.
The van’t Hoff factor is given by i = actual number of particles in solution after dissociation / number of formula units initially
dissolved in solution. Basically, i is given by the number of atoms in the molecule. For example, i in NaCl is two while it is three
in CaCl2. We use this when finding colligative properties of electrolytic solutions. We just add this into the equations, to make
them ∆Tb = iKbm, ∆Tf = iKfm, and π = iMRT. An ion pair is made up of one or more cations and one or more anions held
together by electrostatic forces. Electrolyes containing multicharged ions have a greater tendency to form ion pairs than
electrolytes made up of singly charged ions.
o Calculate the van’t Hoff factor at this concentration: The freezing point of depression of a 0.100 m MgSO4 solution is
0.225 C. First, write out π = iMRT, which we can transform into π / MRT = i. Now, we just plug in our values after
finding them. 0.100 m means 0.100 moles of MgSO4 per kg of solution and the freezing
A colloid is a dispersion of particles of one substance (the dispersed phase) throughout a dispersing medium made of another
substance. One way to distinguish colloids from solutions is by the Tyndall Effect, which scatters light when it is passed through
a colloid.
Hydrophillic is water-loving while Hydrophobic is water-fearing. Colloids are divided into these two categories.
Formula Summary
o molality (m) = moles of solute / moles of solvent (kg)
o c = kP
o P1 = X1P1o
o ∆P = X2P1o
o ∆Tb = Kbm
o ∆Tf = Kfm
o π = MRT
o i= actual # particles in soln after dissociation / number of formula units initially dissolved in soln